Solve for the rational numbers x and y: \(2^{x+y} \cdot 3^{x-y} \cdot 6^{2x+2y}= 72.\)

Guest Jul 13, 2019

#2**+5 **

\(2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2x+2y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,6^{2(x+y)}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,(6^2)^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,3^{x-y}\,\cdot\,36^{x+y}\ =\ 72\\~\\ 2^{x+y}\,\cdot\,36^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ (2\,\cdot\,36)^{x+y}\,\cdot\,3^{x-y}\ =\ 72\\~\\ 72^{x+y}\,\cdot\,3^{x-y}\ =\ 72\)

At this point we can see that the equation will definitely be satisfied if

x + y = 1 and

x – y = 0

So x = 1/2 and y = 1/2 is definitely a solution.

But that doesn't necessarily mean it is the only solution.

I don't know how to prove that it is the only rational solution, but it does seem to be.

Maybe someone else can explain why that is.

hectictar Jul 13, 2019

#3**0 **

LHS:

\(\displaystyle 2^{x+y}.3^{x-y}.6^{2x+2y}=2^{x+y}.3^{x-y}.(2\times3)^{2x+2y}=2^{x+y}.3^{x-y}.2^{2x+2y}.3^{2x+2y}\\=2^{3x+3y}.3^{3x+y}\)

RHS:

\(\displaystyle 72=2^{3}.3^{2}\)

It's an identity, so equating indices,

\(\displaystyle 3x+3y=3,\\3x+y=2.\)

Solve, x = 1/2, y = 1/2.

The solution is unique.

Guest Jul 14, 2019